# To prove an inequality regarding to the resolvent of a self-adjoint operator.

I've been studying functional analysis and currently solving problems in "Mathematcial Methods in Quantum Mechanics With Applications to Schrodinger Operators" written by G. Teschl, but I have a difficulty in proving the following problem(Problem 3.7 in the book).

Show that for a self-adjoint operator $$A$$ we have $$\Vert AR_A(z)\Vert \le \frac{|z|}{|Im(z)|},$$ where $$R_A(z) = (A-z)^{-1}$$ is the resolvent.

It seems radical to me. I do not know where to start. I tried $$AR_A(z) = (z+A-z)R_A(z) = zR_A(z) + I$$, but it did not proceed further. I guess the resolvent formula $$R_A(z) - R_A(z') = (z-z')R_A(z)R_A(z')$$ will play the significant role in solving the problem.

Can someone give me a help? Any idea or hint would be aprreciated a lot.

## 1 Answer

I cannot think of a pretty way to do it (fairly sure there has to be). But, you now that $$AR_A(z)$$ is normal. So $$\|AR_A(z)\|=\max\{|\lambda|:\ \lambda\in \sigma(AR_A(z))\}=\max\left\{\frac{|\lambda|}{|\lambda-z|}:\ \lambda\in\sigma(A)\right\}.$$ Since $$A$$ is selfadjoint, we know that $$\sigma(A)\subset\mathbb R$$. Write $$z=a+ib$$. For your formula to work we need $$b\ne0$$. Thus we want to maximize $$\frac{|t|}{|t-a+ib|}=\frac{|t|}{\sqrt{(t-a)^2+b^2}}.$$ Consider the square, we want to maximize the function $$t\longmapsto \frac{t^2}{(t-a)^2+b^2}.$$ Differentiating and equating to zero gives you that the above function has critical points at $$t=0$$ and $$t=(a^2+b^2)/a$$. So $$\|AR_A(z)\|\leq\sqrt{\frac{(a^2+b^2)^2/a^2}{((a^2+b^2)/a-a)^2+b^2}}=\sqrt{\frac{a^2+b^2}{b^2}}=\frac{|z|}{|\operatorname{Im} z|}$$

• This is insane. Surprised by your insight and computation. Thanks Argerami. – Euduardo Dec 9 '18 at 5:43